Let's solve the two questions one by one.

(a) Finding the average rate of change from 0 minutes to 20 minutes

The average rate of change of temperature $R(t)$ over a time interval $[t_1, t_2]$ is calculated by: $\text{Average rate of change} = \frac{R(t_2) - R(t_1)}{t_2 - t_1}$

For the interval from 0 to 20 minutes:

• $t_1 = 0$ and $R(t_1) = 184.3$
• $t_2 = 20$ and $R(t_2) = 140.3$

Substitute these values: $\text{Average rate of change} = \frac{140.3 - 184.3}{20 - 0} = \frac{-44}{20} = -2.2 \, \text{°C per minute}$

So, the answer for part (a) is -2.2 °C per minute.

(b) Finding the average rate of change from 30 minutes to 60 minutes

For the interval from 30 to 60 minutes:

• $t_1 = 30$ and $R(t_1) = 109.3$
• $t_2 = 60$ and $R(t_2) = 28.3$

Substitute these values: $\text{Average rate of change} = \frac{28.3 - 109.3}{60 - 30} = \frac{-81}{30} = -2.7 \, \text{°C per minute}$

So, the answer for part (b) is -2.7 °C per minute.

Would you like a detailed breakdown or have any questions about this? Here are some related questions:

1. What is the significance of a negative rate of change in temperature?
2. How would the average rate of change differ if calculated over a longer time interval?
3. How can the rate of temperature decrease be used to predict cooling time?
4. What mathematical model might describe the cooling of a roast?
5. How would you find the instantaneous rate of change of temperature at a specific time?

Tip: The average rate of change gives a general trend, while the instantaneous rate of change gives precise information at a particular moment.